(A) The volume of a cylinder is given by $V = \pi r^2 h$. By differentiating with respect to time $t$,we get: $\frac{dV}{dt} = \pi \left( 2rh \frac{dr

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(A) The volume of a cylinder is given by $V = \pi r^2 h$. By differentiating with respect to time $t$,we get: $\frac{dV}{dt} = \pi \left( 2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} \right)$. Given: $r = 4 \, m$,$h = 6 \, m$,$\frac{dr}{dt} = 3 \, m/sec$,and $\frac{dh}{dt} = -4 \, m/sec$ (since altitude is decreasing). Substituting these values: $\frac{dV}{dt} = \pi \left[ 2(4)(6)(3) + (4)^2(-4) \right]$. $\frac{dV}{dt} = \pi [144 - 64] = 80\pi \, m^3/sec$.

Class 12 Mathematics Application of Derivatives Rate of Change of Quantities

Medium

The radius of a cylinder is increasing at the rate of $3 \, m/sec$ and its altitude is decreasing at the rate of $4 \, m/sec$. The rate of change of volume when the radius is $4 \, m$ and the altitude is $6 \, m$ is:

A
$80\pi \, m^3/sec$
B
$144\pi \, m^3/sec$
C
$80 \, m^3/sec$
D
$64 \, m^3/sec$

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Application of Derivatives

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Rate of Change of Quantities

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The radius of a cylinder is increasing at the rate of $3 \, m/sec$ and its altitude is decreasing at the rate of $4 \, m/sec$. The rate of change of volume when the radius is $4 \, m$ and the altitude is $6 \, m$ is:

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